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### pH question

Posted: Mon Jan 14, 2019 10:21 pm
On 6B.3 part B (7th ed) it asks what the actual pH of a solution would be when a lab technician wants to prepare 200 mL of a 0.025 M HCL solution but uses a volumetric flask of 250 mL by mistake. The solutions show that you take the -log((200mL x 0.025M)/250 mL)= 1.7. Can someone explain this work or the logic behind it? I understand that we take the negative log of the concentration of H3O (or H+) but why (200x.025)/250?

### Re: pH question

Posted: Mon Jan 14, 2019 10:26 pm
You need to use the formula M1V1=M2V2 because the volume of the container was changed. You have the original volume and concentration, and the new volume, so you must solve for the new concentration.

### Re: pH question

Posted: Tue Jan 15, 2019 7:15 pm
I have the 6th edition textbook and solution manual for it. As I was working on 12.29 part d, i noticed that I got pOH=0.15 & pH=13.85 but the solutions manual has pOH=3.15 and pH=10.85 . The way i solved was ((2.00 mL/.500mL)*(.175))= .7 then i did -log(.7) which gave me the pOH=.15

what the book did ((2.00mL/500mL)*(175))= 7.0 *10^-4 -log(7.0*10^-4)

can someone please explain to me why the solutions manual used 500mL instead of .500mL?

### Re: pH question

Posted: Tue Jan 15, 2019 7:21 pm
504909207 wrote:I have the 6th edition textbook and solution manual for it. As I was working on 12.29 part d, i noticed that I got pOH=0.15 & pH=13.85 but the solutions manual has pOH=3.15 and pH=10.85 . The way i solved was ((2.00 mL/.500mL)*(.175))= .7 then i did -log(.7) which gave me the pOH=.15

what the book did ((2.00mL/500mL)*(175))= 7.0 *10^-4 -log(7.0*10^-4)

can someone please explain to me why the solutions manual used 500mL instead of .500mL?

The question says, " 2.00 mL of 0.175 m KOH(aq) after dilution to 0.500 L," so 0.500 L is equivalent to 500 mL, not 0.500 mL.